$$ \left\{\begin{array}{l} ext { QUESTION } 3 \ \left.\begin{array}{ll}3.1 ext { Calculate: } & \sum_{n=1}^{\infty} \frac{27}{10^{2 n}} \ 3.2 & 1+2+4+\ldots ext { is a geometric series. } \ 3.2 .1 ext { Write down the }(n+1)^{t h} ext { term. } \ 3.2 .2 ext { Calculate the sum of } n ext { terms. } \ 3.2 .3 ext { Hence, give the relation between the sum of } 20 ext { terms and the } 21^{ ext {st }} ext { term. (2) } \ ext { [10] }\end{array} ight]\end{array} ight. $$

Question

$$ \left\{\begin{array}{l}	ext { QUESTION } 3 \ \left.\begin{array}{ll}3.1 	ext { Calculate: } & \sum_{n=1}^{\infty} \frac{27}{10^{2 n}} \ 3.2 & 1+2+4+\ldots 	ext { is a geometric series. } \ 3.2 .1 	ext { Write down the }(n+1)^{t h} 	ext { term. } \ 3.2 .2 	ext { Calculate the sum of } n 	ext { terms. } \ 3.2 .3 	ext { Hence, give the relation between the sum of } 20 	ext { terms and the } 21^{	ext {st }} 	ext { term. (2) } \ 	ext { [10] }\end{array}ight]\end{array}ight. $$

UpStudy AI · Accepted Answer

Certainly! Let's address each part of **Question 3** step by step.

---

### **Question 3.1**

**Calculate:**  
$$
\sum_{n=1}^{\infty} \frac{27}{10^{2n}}
$$

**Solution:**

This is an infinite geometric series of the form:
$$
\sum_{n=1}^{\infty} ar^{n-1}
$$
where:
- $$ a $$ is the first term
- $$ r $$ is the common ratio

However, our series is written as:
$$
\sum_{n=1}^{\infty} \frac{27}{10^{2n}} = 27 \sum_{n=1}^{\infty} \left(\frac{1}{100}\right)^n
$$
Here, the first term $$ a = \frac{27}{10^{2}} = \frac{27}{100} $$ and the common ratio $$ r = \frac{1}{100} = 0.01 $$.

The sum $$ S $$ of an infinite geometric series where $$ |r| < 1 $$ is:
$$
S = \frac{a}{1 - r}
$$

Plugging in the values:
$$
S = \frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{\frac{27}{100}}{\frac{99}{100}} = \frac{27}{99} = \frac{9}{33} = \frac{3}{11}
$$

**Answer:**  
$$
\sum_{n=1}^{\infty} \frac{27}{10^{2n}} = \frac{3}{11}
$$

---

### **Question 3.2**

**Given:**  
The series $$ 1 + 2 + 4 + \ldots $$ is a geometric series.

#### **3.2.1 Write down the $$(n+1)^{\text{th}}$$ term.**

**Solution:**

A geometric series has the general form:
$$
a, \, ar, \, ar^2, \, ar^3, \, \ldots
$$
where:
- $$ a $$ is the first term
- $$ r $$ is the common ratio

For the given series:
$$
1 + 2 + 4 + \ldots
$$
- First term $$ a = 1 $$
- Common ratio $$ r = 2 $$ (since each term is multiplied by 2 to get the next term)

The $$ n^{\text{th}} $$ term is:
$$
a_n = ar^{n-1} = 1 \times 2^{n-1} = 2^{n-1}
$$

Thus, the $$(n+1)^{\text{th}}$$ term is:
$$
a_{n+1} = 2^{(n+1)-1} = 2^{n}
$$

**Answer:**  
The $$(n+1)^{\text{th}}$$ term is $$ 2^{n} $$.

#### **3.2.2 Calculate the sum of $$ n $$ terms.**

**Solution:**

The sum $$ S_n $$ of the first $$ n $$ terms of a geometric series is given by:
$$
S_n = a \frac{r^n - 1}{r - 1}
$$
where:
- $$ a = 1 $$
- $$ r = 2 $$

Plugging in the values:
$$
S_n = 1 \times \frac{2^n - 1}{2 - 1} = 2^n - 1
$$

**Answer:**  
The sum of the first $$ n $$ terms is $$ S_n = 2^{n} - 1 $$.

#### **3.2.3 Hence, give the relation between the sum of 20 terms and the 21ᵗʰ term.**

**Solution:**

From **3.2.2**, the sum of the first $$ n $$ terms is:
$$
S_n = 2^n - 1
$$
And from **3.2.1**, the $$(n+1)^{\text{th}}$$ term is:
$$
a_{n+1} = 2^{n}
$$

Let's set $$ n = 20 $$:

- Sum of the first 20 terms:
$$
S_{20} = 2^{20} - 1
$$
- $$ 21^{\text{st}} $$ term:
$$
a_{21} = 2^{20}
$$

Notice that:
$$
S_{20} = a_{21} - 1
$$
$$
\Rightarrow S_{20} + 1 = a_{21}
$$

**Relation:**  
$$
\text{Sum of 20 terms} + 1 = \text{21ᵗʰ term}
$$
Or,
$$
S_{20} = a_{21} - 1
$$

**Answer:**  
The sum of the first 20 terms is equal to the 21ᵗʰ term minus one, that is:
$$
S_{20} = a_{21} - 1
$$

---

### **Summary of Answers**

1. **3.1:**  
   $$
   \sum_{n=1}^{\infty} \frac{27}{10^{2n}} = \frac{3}{11}
   $$

2. **3.2.1:**  
   The $$(n+1)^{\text{th}}$$ term is $$ 2^{n} $$.

3. **3.2.2:**  
   The sum of the first $$ n $$ terms is $$ S_n = 2^{n} - 1 $$.

4. **3.2.3:**  
   The sum of the first 20 terms is equal to the 21ᵗʰ term minus one:
   $$
   S_{20} = a_{21} - 1
   $$

---
**Question 3 Solutions:** 1. **3.1:** $$ \sum_{n=1}^{\infty} \frac{27}{10^{2n}} = \frac{3}{11} $$ 2. **3.2.1:** The $$(n+1)^{\text{th}}$$ term is $$ 2^{n} $$. 3. **3.2.2:** The sum of the first $$ n $$ terms is $$ S_n = 2^{n} - 1 $$. 4. **3.2.3:** The sum of the first 20 terms is equal to the 21ᵗʰ term minus one: $$ S_{20} = a_{21} - 1 $$

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